The remainder obtained when 999⁸ is divided by 19 is

(A) 13

(B) 11

(C) 9

(D) 7

The remainder obtained when 999⁸ is divided by 19 is

(A) 13

(B) 11

(C) 9

(D) 7

D. 7

I used the calculator to divide 999 by 19 and found remainder 11. Then I found the remainder of 999 x 999 divided by 19 and got 7. This will be the repeating pattern for all odd and even powers of 999 because the units digit is 9.

8 is even, so 999⁸ divided by 19 will have a remainder of 7

Edit: the explanation is shady, please forgive me if it gave you a headache. I ran it through wolframalpha to correct my faulty reasoning. got lucky with the answer though.

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999^8 \div19=(988+11)^8 , \text{now for even power of 11 the remainder will be 7}

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But will it have repeating pattern? I mean my way to approach the problem was to focus on the last two digit of the product. In fact, if we do 999³ - 11 then it’s not divisible by 19. So it seems pattern is not repeating for remainder. I am not getting the right way to approach the problem. Obviously we have to focus on the 11⁸ ( because we can write 999 as (988+11)⁸ as 988 is divisible by 19 so we have to focus on 11⁸ for remainder). Consequently, we have to try all last two digits of products to get the remainder or patter of remainder ?

Yeah so the last digit will be always 1 or we have to focus on the last two digit of the product ? Like 21, 31 etc but we can’t tell which will end up as even one and that too on the 8 th power.

the last digit will always be 1 in even powers of 11.

Yeah, but my analysis was kind of wonky. Not only was I confidently incorrect, I also got astronomically lucky. Here is how I atone for it, please forgive me.

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Update

I solved it somehow it was really full of traps and approximation

So we end up with 11⁸ / 19

Now key point here is to use Pascal triangle

We can actually calculate 11⁸ way faster with help of Pascal triangle.

So we know like ending two digit of 11² = 21

For 11³ is 31

For 11⁴ is 41

Similarly last two digit of 11⁸ will be 81

But thing is we can’t rely on last two digit when we divide by double digit number in this case 19

So we need at least last 3 digits

Then I calculated the whole value of 11⁸ with help of Pascal triangle and the ending 3 digits were 881.

So when we divide 881 by 19 remainder is 7.

This question was full of trap like we can’t completely rely on unit digit for remainder when we divide the number with two digit number specially.

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Could somebody point me to a video where greg discusses the concepts necessary to solve this kind of questions containing remainders?